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Below we have a system A and B in thermal equilibrium. What happens to the entropy and energy of the systems when you inject heat into system A and let it equilibriate further? I know that heat will flow from A to B, the total entropy will increase, and temperature of the systems will increase. However, will points representing the new entropy and energy combination of system A and B stay on the same curves shown in figure 6.4? Or will we have to redraw new curves and move the points to a location that is not on the original curves? Thank you.

From Molecular Driving Forces textbook

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If all other properties of the two systems and the combined system are the same, and only the temperature increases, why would they leave the curve? Only the slope will decrease with increasing T. The slope 1/T of system B will slowly follow that of A until they become parallel again (equilibrium reached again).

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  • $\begingroup$ I see, what could be some scenarios (other than increasing volume and introducing more particles) where the points would leave the curve? $\endgroup$
    – Stop5
    Sep 22 at 0:25
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You have not specified so but let us assume that the two parts, $A$ and $B$, after the injection of the thermal energy into $A$, are isolated together and their only interaction is with each other via conduction.

As a result of the injection of thermal energy into $A$ both its internal energy $U_A$ and its entropy $S_A$ will increase; the monotonically increasing curve representing this is the lower plot on the figure. At start the tangent will be closer to horizontal corresponding to an increase of temperature by the added thermal energy and no work.

If now you let them equilibrate via conduction then $A$ starts at a higher temperature and $B$ at a lower one. Since they are isolated together, their equilibrium is reached when their total entropy is maximum while the sum of their individual internal energies is held constant. At that point, "state", the individual temperatures will be the same, that is, the tangents of the curves will be parallel again. The slope of the common temperature line will be between the one shown on the figure and another one corresponding to the higher starting temperature of $A$.

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